F

对于每一棵子树，考虑如何快速进行配对，ma为子树中权值最大的颜色的权值，sum为子树总权值，可以发现，如果ma>sum/2，那么这棵子树的答案就是sum-ma,否则答案就是sum/2。维护子树信息可以通过线段树合并或dsu on tree

线段树合并代码

#include <bits/stdc++.h>
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))

namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
    bool IOerror=0;
    inline char nc() {
        static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
        if(p1==pend) {
            p1=buf;
            pend=buf+fread(buf,1,BUF_SIZE,stdin);
            if(pend==p1) {
                IOerror=1;
                return -1;
            }
        }
        return *p1++;
    }
    inline bool blank(char ch) {
        return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
    }
    template<class T> inline bool read(T &x) {
        bool sign=0;
        char ch=nc();
        x=0;
        for(; blank(ch); ch=nc());
        if(IOerror)return false;
        if(ch=='-')sign=1,ch=nc();
        for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
        if(sign)x=-x;
        return true;
    }
    template<class T,class... U>bool read(T& h,U&... t) {
        return read(h)&&read(t...);
    }
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int INF = 0x3f3f3f3f;
const int N = 2e5+10;

vector<int>e[N];
int ans[N],res;

struct node {
    int l,r,ma;
} tree[N * 40];

int a[N],w[N],siz[N];

int root[N],cnt;
#define ls tree[x].l
#define rs tree[x].r
void ins(int &x,int l,int r,int pos,int val) {
    if(!x) x=++cnt;
    if(l==r) {
        tree[x].ma+=val;
        return ;
    }
    int mid=l+r>>1;
    if(pos<=mid) ins(ls,l,mid,pos,val);
    else ins(rs,mid+1,r,pos,val);
    tree[x].ma=max(tree[ls].ma,tree[rs].ma);
}

int merge(int x,int y,int l,int r) {
    if(!x || !y) return x+y;
    if(l==r) {
        tree[x].ma+=tree[y].ma;
        return x;
    }
    int mid=l+r>>1;
    tree[x].l=merge(tree[x].l,tree[y].l,l,mid);
    tree[x].r=merge(tree[x].r,tree[y].r,mid+1,r);
    tree[x].ma=max(tree[ls].ma,tree[rs].ma);
    return x;
}


void dfs(int u,int fa) {
    siz[u]=w[u];
    ins(root[u],1,N,a[u],w[u]);
    for(auto v:e[u]) {
        if(v==fa) continue;
        dfs(v,u);
        siz[u]+=siz[v];
        root[u]=merge(root[u],root[v],1,N);
    }
    int tot=tree[root[u]].ma;
    if(tot*2>=siz[u]) ans[u]=siz[u]-tot;
    else ans[u]=siz[u]/2;
}

int main() {

#ifdef xiaofan
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
#endif

    int n;
    read(n);
    for(int i=1; i<n; i++) {
        int u,v;
        read(u,v);
        e[u].pb(v);
        e[v].pb(u);
    }
    for(int i=1; i<=n; i++) read(a[i],w[i]);
    dfs(1,0);
    for(int i=1; i<=n; i++) printf("%d\n",ans[i]);



    return 0;
}